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# (Solved): For a square matrix A, the statement (A-lambda I)v=0hasanon-zerosolutionv is equivalent to the sta ...

For a square matrix

A

, the statement

 (A-lambda I)v=0hasanon-zerosolutionv

is equivalent to the statement

:'det(A-\lambda I)=0 :

The quantity

det(A-\lambda I)

is a polynomial in

\lambda

known as the characteristic polynomial of

A

and its roots are the eigenvalues of

A

. This gives us a technique for finding the eigenvalues of

A

Let's check our understanding. i) The characteristic polynomial of

A=([1,2],[2,1])

is

det(A-tI)=det([1-t,2],[2,1-t])=(1-t)^(2)-4,0 a

Ordered

t_(1)=◻t_(2)=◻◻v_(1)=◻v_(2)=◻((1)/(2))(:1,2:)B=([5,-1,0],[-1,5,0],[0,0,-3])det(B-tI)=◻Bt_(1)=-3,t_(2)=◻,t_(3)=◻ 目, t_(1) this has roots (and hence B has eigenvalues) t_(1)=-3,t_(2)=◻,t_(3)=◻ 目, t_(1) the roots of this polynomial are t_(1)= ◻ and t_(2)= ◻ Ordered the same way, the associated eigenvectors are ◻ v_(1)= ◻ and v_(2)= ◻ Note: the Maple notation for the vector ((1)/(2)) is (:1,2:) ii

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